Timeline for Witten's QFT and Jones Poly paper
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2011 at 0:13 | comment | added | Kevin Wray | @Paul: Yes, I realized this last night. Everything is ok now. Thanks for the explanation. | |
| Feb 22, 2011 at 13:10 | comment | added | Paul | This is a homotopy obstruction, not an extension obstruction. Of course they are related: Think of it as a relative extension problem for $(M\times I, M\times\{0,1\})$, then use the suspension iso $H^i(M\times I, M\times\{0,1\})=H^{i-1}(M)$. This explains the shift. | |
| Feb 22, 2011 at 1:30 | comment | added | Kevin Wray | Ok, this I like! For some reason though I was thinking that the obstruction (extending over the $3$-skeleton) was an element in $H^3\big(M;\pi_2(G)\big)$, not in $H^3\big(M;\pi_3(G)\big)$. Maybe that is the obstruction to extending a section over the $G$-bundle? | |
| Feb 22, 2011 at 0:55 | comment | added | Paul | For any (simply connected) space $G$, the obstructions to nulhomotoping $f:M\to G$ lie in $H^i(M;\pi_i(G))$. If $\pi_1(G)=0=\pi_2(G)$, then since $M$ is 3-dimensional there is only one obstruction in $H^3(M;\pi_3(G))$. So $\pi_3(G)=Z$ means the only (primary) obstruction is in $H^3(M;Z)$. Although I'm citing obstruction theory, this is elementary since you can easily nullhomotop the 2-skeleton of $M$, so the map factors through $M/2-skeleton=S^3$ for an appropriate cell structure. Incidentally, for $G=Spin(4)$, $\pi_3(G)=Z\oplus Z$ since $Spin(4)=SU(2)\times SU(2)$. | |
| Feb 22, 2011 at 0:40 | comment | added | Kevin Wray | Yes, I agree with what everyone has said. I am just trying to understand Witten's reasoning when he says that since $\pi_3(G) \cong \mathbb{Z})$ (i.e., when $\pi_3(G)$ is non-trivial) we have nontrivial gauge transformations. I understand Konrad's description (since $\pi_1(G)=\pi_2(G) =0$ there is no obstruction to a global section - giving gauge transformations as maps $M\rightarrow G$), but he is starting with a cohomological statement. So, is it that Witten is just thinking about the Hurewicz isom. to got to homology, then cohom., or is it straightforward from $pi_3(G)\neq 0$? | |
| Feb 22, 2011 at 0:26 | history | answered | Paul | CC BY-SA 2.5 |